Voltage Null Points of a Mechanical Meter Bridge

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The discussion centers on the concept of voltage null points in a mechanical meter bridge, focusing on the relationship between wire resistance and potential difference. It is argued that for point D to be a null point, the resistance of segments AD and AB must be equal. Moving the jockey to the left reduces the resistance of AD, increasing the potential at D compared to B, which should theoretically cause current to flow from D to B. However, the textbook states that when the jockey is to the left of D, no current flows at point B, leading to confusion. Clarification is sought on why the textbook's assertion contradicts the reasoning presented.
Shreya
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Homework Statement
The answers of this question is mentioned to be A & C. My question is why isn't B true?
Please be kind to help
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My reasoning is that the wire has a resistance/unit length. The resistance of AD must equal that of AB for D to be called a null point. If we move jockey to the left, the resistance of AD decreases, thereby decreasing the potential drop across it. This means that the V of D is more than that of B causing a current to flow from D to B. So according to me B should be an answer, while the textbook says.
 

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when jockey is connect to the left of D no current in the wire B. that is why its wrong
 
salmanshu322 said:
when jockey is connect to the left of D no current in the wire B. that is why its wrong
I cannot understand why it is so. Could you please explain? According to my reasoning the resistance of AD< that of AB. Therefore, the V drop across AD is lower making D at a higher potential than B, causing a current to flow from D to B.
 

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